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I am asked to find the equation of a cubic function that passes through the origin. It also passes through the points $(1, 3), (2, 6),$ and $(-1, 10)$.

I have walked through many answers for similar questions that suggest to use a substitution method by subbing in all the points and writing in terms of variables. I have tried that but I don't really know where to take it from there or what variables to write it as.

If anyone could provide their working out for this problem it would be extremely enlightening.

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the general cubic equation is $$y=ax^3+bx^2+cx+d.$$Plug in the coordinates of the points for x and y, and you end up with a system of four equations in four variables, namely $a, b, c$ and $d$. Hope that helps!

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  • $\begingroup$ I have done as you said and I have ended up with an equation for a=, b=, c=, and d=0(given). Where should I go next? The there are two variables on the RHS of these equations and I don't know how to proceed further. $\endgroup$ – Haotian Huang Feb 18 '18 at 7:13
  • $\begingroup$ Use row operations to isolate the variables. For example, get an expression for a by isolating it and substitute it into another, until you get equations in one variable.(Or, if you know how, set up a matrix and row-reduce it.) $\endgroup$ – Ranjeev Grewal Feb 18 '18 at 7:17
  • $\begingroup$ I've finally done it! Thank you very much Ranjeev $\endgroup$ – Haotian Huang Feb 18 '18 at 7:55
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Given four points $(x_i,y_i)$ consider the functions $$f_1(x)=\frac {(x-x_2)(x-x_3)(x-x_4)}{(x_1-x_2)(x_1-x_3)(x_1-x_4)}$$ so that $f_1(x_1)=1$ and $f_1(x_i)=0, i\neq 1$, and similarly $f_2, f_3, f_4$. Note that the $f_i$ are cubic in $x$.

Then $p(x)=y_1f_1(x)+y_2f_2(x)+y_3f_3(x)+y_4f_4(x)$ is at most a cubic polynomial and passes through the four given points.

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  • $\begingroup$ See Lagrange Interpolation/Polynomials en.wikipedia.org/wiki/Lagrange_polynomial Note also that this is not the only place where a system with $f_i(x_j)=1(i=j) =0 (i\neq j)$ is useful. See for example (versions of) the Chinese Remainder Theorem. $\endgroup$ – Mark Bennet Feb 18 '18 at 15:01
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Guide:

Let the equation be $y=ax^3+bx^2+cx+d$, since it passes through $(1,3)$, we have

$$a(1)^3+b(1)^2+c(1)+d=3$$

Do the same thing for the other $3$ points.

Hence you will obtain $4$ linear equation with $4$ variables.

You can then solve it using elementary row operations to recover $a,b,c,d$.

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  • $\begingroup$ Unfortunately, I have not learn elementary row operations yet. However I have obtained 4 linear equations with 4 variables. Is there a way to find the cubic equation without elementary row operations? $\endgroup$ – Haotian Huang Feb 18 '18 at 7:16
  • $\begingroup$ what about substitution? $\endgroup$ – Siong Thye Goh Feb 18 '18 at 7:16
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To determine a conic we need to solve 5 equations with 5 given points.

Likewise here we are given 4 points and 4 simultaneous equations, not involving any $xy $ term.. so solve it by Cramer's determinants.

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An equation of the cubic that passes through the four points $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$ and $(x_4,y_4)$ is $$\begin{vmatrix} x^3 & x^2 & x & y & 1 \\ x_1^3 & x_1^2 & x_1 & y_1 & 1 \\ x_2^3 & x_2^2 & x_2 & y_2 & 1 \\ x_3^3 & x_3^2 & x_3 & y_3 & 1 \\ x_4^3 & x_4^2 & x_4 & y_4 & 1 \end{vmatrix} = 0.$$ Plug in the coordinates of your points and simplify.

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Hint:

It is proper to use the Lagrange function as following$$f(x)=\sum_{cyc}\dfrac{(x-x_2)(x-x_3)(x-x_4)}{(x_1-x_2)(x_1-x_3)(x_1-x_4)}$$

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